International Journal of Fuzzy Systems, Vol. 14, No. 2, June 2012
193
Some Results on Generalized Interval-Valued Intuitionistic Fuzzy Sets Monoranjan Bhowmik and Madhumangal Pal Abstract1 In this paper, we define some results on generalized interval-valued intuitionistic fuzzy sets (GIVIFSs). In fact, all interval-valued intuitionistic fuzzy sets (IVIFSs) are GIVIFSs but all GIVIFSs are not IVIFSs. We define two composite relations, four types of reflexivity and irreflexivity of GIVIFSs with some of their properties. We define generalized interval-valued intuitionistic fuzzy relation (GIVIFR) with some results. Also we define two operators C and I with some properties over GIVIFSs. Finally, an illustrative example is given to using GIVIFSs in a decision making problem. Keywords: operators.
IVIFS, GIVIFS, GIVIFR, C and I
1. Introduction In 1965 , Zadeh [22] introduced the concept of fuzzy subsets. It is well known that fuzzy sets played major role in various areas such as mathematics, physics, statistics, engineering, social sciences and many other fields. Several works on fuzzy sets and applications of fuzzy sets are available in different journals and in books also. Atanassov [6] introduced some operations over interval-valued fuzzy set. Pal and Shyamal [19, 20] introduced fuzzy matrices, interval-valued fuzzy matrices and shown several properties of them. After two decades, Atanassov [4, 7] introduced the concept of intuitionistic fuzzy sets (IFSs), which is a generalization of fuzzy subsets and defines new operations on IFSs [8]. Several authors present a number of results using IFSs. Nanda [15] studied intuitionistic fuzzy relations over IFSs. By the concept of IFSs, first time Pal [16, 17] introduced intuitionistic fuzzy determinant and intuitionistic fuzzy matrices. Latter on Pal et al. [9, 18] Corresponding Author: Monoranjan Bhowmik is with the Department of Mathematics, V.T.T. College, Midnapore, West Bengal-721101, INDIA. E-mail:
[email protected] Madhumangal Pal is with the Department of Applied Mathematics with Oceanology and Computer Programming, Vidyasagar University, Midnapore 721 102, India. E-mail:
[email protected] Manuscript received 8 Nov. 2010; revised 29 Dec. 2011; accepted 10 May 2012.
introduced intuitionistic fuzzy matrices, distance between intuitionistic fuzzy matrices and some properties of intuitionistic fuzzy matrices. Mondal and Samanta [14] gives the idea of generalized IFS. The main concept of this GIFS is – “an element x of X is called significant with respect to a fuzzy subset A of X if the degree of membership μ A (x) > 0.5 , otherwise, it is insignificant.” We see that for a fuzzy subset A both the degrees of membership μ A (x) and
non- membership ν A ( x) = 1 − μ A ( x) can not be significant. Further, we know the definition for an IFS where A = {〈 x, μ A ( x),ν A ( x)〉 / x ∈ X },0 ≤ μ A ( x) + ν A ( x) ≤ 1
for all x ∈ X , so it is observed that μ A ( x) ∧ν A ( x) ≤ 0.5 , for all x ∈ X . Xia [21] induced generalized intuitionistic fuzzy operators. Again Gargo and Atanassov [5] introduced the interval-valued intuitionistic fuzzy sets (IVIFSs) with several properties on IVIFSs and shown applications of IVIFSs. Jana and Pal [13] studied some operators defined over IVIFS. Recently Adak and Bhowmik [1] done interval cut-set of IVIFSs and studied some properties. Using this conception of GIFSs Bhowmik and Pal [10] introduced GIFMS with some properties. Parhaps first time Bhowmik et.al. [11,12] introduce the definition of GIVIFs and studied several properties. Bhowmik et. al. [2] also studied semi ring of interval-valued intuitionistic fuzzy matrices. Lee et. al. [3] give an application on medical diagonosis using IVIFSs. Here, we present two fundamental operators ∨ and ∧ defined over fuzzy sets below: On the interval [0,1] (where the fuzzy sets takes their elements) the following operations are defined for all x, y ∈ [0,1] : x ∨ y = max( x, y ) , x ∧ y = min( x, y). Having motivated from the observation IVIFS, we define GIVIFS as follows: Definition 1 [11]: If the IVIFS A = {〈[ M AL ( x), M AU ( x)], [ N AL ( x), N AU ( x)]〉} , x ∈ X satisfying the condition M AU ( x) ∧ N AU ( x) ≤ 0.5 for all x ∈ X then A is called GIVIFS. The condition M AU ( x) ∧ N AU ( x) ≤ 0.5 is called generalized interval-valued intuitionistic fuzzy condition (GIVIFC). The maximum value of M AU ( x)
and N AU ( x) is 1.0 , therefore GIVIFC implies
© 2012 TFSA
that
International Journal of Fuzzy Systems, Vol. 14, No. 2, June 2012
194
0 ≤ M AU ( x) + N AU ( x) ≤ 1.5 . In this paper, in section 2 we recall some operations on GIVIFSs. In section 3, we have shown by means of example that all the cartesian product of [6] are not valid in form of GIVIFss. In section 4, define different kind of reflexivity and irrflexivity over GIVIFSs with some results. In section 5 and section 6 defines some more results on GIVIFSs and two operators (C and I ) over GIVIFSs with some properties. Finally in section 7 and section 8, we illustrate an example using of GIVIFSs and the coclusion part of this paper.
N BL ( x ) − N AL ( x ).N BL ( x ), N AU ( x ) + N BU ( x ) −
(3)
[
(4)
N
AL
(x) + N 2
BL
(x)
(5)
(6)
,
N
AU
AU
(x) + N 2
( x ) + M BU ( x ) ], 2 BU
(x)
]〉 }
A$B = {〈[ M AL ( x).M BL ( x), M AU ( x).M BU ( x)],
[ N AL ( x ). N BL ( x ) ,
2. Some operations on GIVIFSs Let A and B be two GIVIFSs or IVIFSs on X, where A = {〈[ M AL ( x), M AU ( x)],[ N AL ( x), N AU ( x)]〉},
N AU ( x).N BU ( x)]〉}. A@B M ( x ) + M BL ( x ) M = {〈 x , [ A L , 2
N AU ( x ). N BU ( x ) ]〉}.
2.M AL ( x).M BL ( x) 2.M AU ( x).M BU ( x) , ], M AL ( x) + M BL ( x) M AU ( x) + M BU ( x) 2 . N A L ( x ). N B L ( x ) 2 . N A U ( x ). N B U ( x ) [ , ]〉 } . N AL ( x ) + N BL ( x ) N AU ( x ) + N BU ( x ) A#B = {〈[
n [n]A ={〈[1− (1− M AL ( x))n ,1− (1− M AU (x))n ],[ NAL ( x),.
n N AU ( x )]〉}, n > 0 .
n n B = {〈[ M BL ( x), M BU ( x)],[ N BL ( x), N BU ( x)]〉} . (7) A[ n ] = {〈[ M AL ( x), M AU ( x)],[1 − (1 − N AL ( x)) n ,1 − In the following we define some relational operations (1 − N AU ( x )) n ]〉 /}, n > 0, on GIVIFSs. We have shown in [11] that all of them are valid for (1) A = B ⇔{MAL (x) = MBL (x) and NAU (x) = NBU (x)}, for IVIFSs but not valid for GIVIFSs. all X. (2) A ≺ B ⇔{MAL (x) ≤ MBL (x) and MAU (x) ≤ MBU (x)}, 3. Cartesian product of GIVIFSs
{N AL ( x) ≥ N BL ( x) and N AU ( x) ≥ N BU ( x)},∀x. (3)
all X. (4)
A ⊂◊ B iff {NAL (x) ≥ NBL (x) and NAU (x) ≥ NBU (x)}, for all X.
(5)
Here we recall five cartesian products from [6] and we
A⊂Δ Biff {MAL(x) ≤ MBL(x)and MAU(x) ≤ MBU(x)}, for shall prove that the results of all of them do not form
A⊆Biff {(MAU(x) ≤MBU(x)and MAL(x) ≤MBL(x))}and {(NAU (x) ≥ NBU (x) and NAL (x) ≥ NBL (x))},for all x∈ X
Here, we define some unary and binary operations on GIVIFSs: (1) A = {〈 x, N A ( x), M A ( x)〉 / x ∈ X }, for all x ∈ X . (2) A∩B ={〈[min(MAL (x), MBL (x)),min(MAU (x), MBU (x))], [max(NAL (x), NBL(x)),max(NAU (x), NBU (x))]〉 / x∈X}. (3) A∪B={[m 〈 ax(MAL(x),MBL(x)),max(MAU(x),MBU(x))], [min (NAL (x), NBL (x)),min (NAU (x), NBU (x))]〉 / x∈ X}. The subsets A , A ∩ B and A ∪ B all are GIVIFSs. Now, we recall some operations for IVIFSs [5, 13]. Definition 2 [13]: For any two IVIFSs A and B on X, (1) A+ B ={〈[MAL (x) + MBL (x) − MAL (x).MBL (x), MAU (x) + MBU ( x) − MBU ( x) − M AU ( x).MBU ( x)],[ NAL (x).NBL ( x), N AU ( x).N BU ( x)]〉. (2) A.B = {〈[ M AL ( x).M BL ( x), M AU ( x).M BU ( x)],[ N AL ( x) +
GIVIFSs. Definition 3 [6]: Let X , Y be two universes and also let A ∈ X and B ∈ Y be two GIVIFSs. The cartesian product A and B are defined as follows: (1) A × 1 B = { 〈[ m in{ M A L ( x ), M B L ( y )}, min { M AU ( x ), M BU ( y )}],[ max { N AL ( x ), N BL ( y )}, max{N AU ( x), N BU ( y )}]〉 / x ∈ X , y ∈ Y }. (2) A × 2 B = {〈[ max{ M AL ( x ), M BL ( y )}, max {M AU ( x ), M BU ( y )}],[ min { N AL ( x ), N BL ( y )}, min{N AU ( x), N BU ( y )}]〉 / x ∈ X , y ∈ Y }. (3) A ×3 B = {〈[ M AL ( x).M BL ( y ), M AU ( x).M BU ( y )], [ N AL ( x).N BL ( y ), N AU ( x ).N BU ( y )]〉 / x ∈ X , y ∈ Y }. (4) A ×4 B = {〈[ M AL ( x) + M BL ( y ) − M AL ( x).M BL ( y ), M AU ( x) + M BU ( y ) − M AU ( x).M BU ( y )], [ N AL ( x).N BL ( y ), N AU ( x).N BU ( y )]〉 /x ∈ X , y ∈ Y }. (5) A ×5 B = {〈[ M AL ( x).M BL ( y ), M AU ( x).M BU ( y )], [ N AL ( x) + N BL ( y ) − N AL ( x).N BL ( y ), N AU ( x) + N BU ( x) − N AU ( y ).N BU ( y )]〉 /x ∈ X , y ∈ Y }. We have shown by means of examples that ×4 ,×5 are not GIVIFSs. Example 1: Let A and B be two GIVIFSs on X ,
M. Bhowmik et al.: Some Results on Generalized Interval-Valued Intuitionistic Fuzzy Sets
A = {〈x1 , [0.3,0.4], [0.6,0.8]〉} and B = {〈 x1 , [0.2, 0.4],[0.5, 0.8]〉} then A ×4 B = {〈 x1 , [0.44,0.64], [0.3.0.64]〉}.
M (P
where
that A, B are GIVIFSs but A ×4 B is not GIVIFS. Example 2: Let A and B be two GIVIFSs on X , where A = {〈x1 , [0.6,0.8], [0.3,0.4]〉} and B = {〈 x1 ,
N(P
, M ( A×
4 B )U
numbers
a p , b p , cq , d q ∈ [0,1],
where
p∈P
p∈P
q∈Q
min (b p .d q ) = ( min b p ).( min d q ).
p∈P , q∈Q
p∈P
q∈Q
y∈Y
x∈ X
= min {max {N RU ( x, y ), N RU ( y , z )}}. y∈Y
y∈Y
x∈ X
x∈ X
x∈X
N ( P∗R ) L = max {min {N RL ( x, y ), N RL ( y , z )}} ; y∈Y
x∈X
N ( P∗R )U = max {min {N RL ( x, y ), N RL ( y, z )}}. y∈Y
x∈ X
Obviously P R and P ∗ R are GIVIFSs.
and
max ( a p .cq ) = ( max a p ).( max cq ),
R )U
= min {max {N RL ( x, y ), N RL ( y , z )}} ;
y∈Y
4. Reflexivity and irreflexivity of GIVIFSs
q ∈ Q , then the following equations hold good p∈P , q∈Q
R)L
x∈X
M ( P∗R )U = min {max {M RU ( x, y ), M RU ( y , z )}}.
5
not a GIVIFS. Proposition 1: Given two index sets P and Q and
x∈ X
M ( P∗R ) L = min {max {M RL ( x, y ), M RL ( y, z )}} ;
[0.5, 0.8],[0.2, 0.4]〉} then A ×5 B = {〈 x1 , [0.3,0.64], [0.44.0.64]〉}. Hence, min{M ( A× B )U , N ( A× B )U } ≤/ 0.5 i.e., A ×5 B is 5
y∈Y
y∈Y
N(P
4 B )U
= max {min {M RL ( x, y ), M RL ( y , z )}} ;
M ( P R )U = max {min {M RU ( x, y ), M RU ( y, z )}}.
} ≤/ 0.5 . It may be noted
Hence, min{M ( A×
R) L
195
In this section we define different kind of reflexivity and irreflexivity of GIVIFSs with some of their results. Definition 7: Let R ∈ GR( X × X ) , then
T1 : R is reflexive of type-1 if M R = [1,1], N R = [0, 0] and from Proposition 1 it is obvious that A ×3 B is a (2) T2 : R is reflexive of type-2 if M R = [1,1] and GIVIFS. N RU ( x, x) ∨ N RU ( y, y ) ≤ N RU ( x, y ) for all Throughout this paper we will use × for ×1 . x, y ∈ X . Definition 4: A generalized interval-valued intuitionistic (3) T : R is reflexive of type-3 if M ( x, x) ∧ 3 RU fuzzy relation (GIVIFR) is a generalized interval-valued M RU ( y, y ) ≥ 0.5 ∧ M RU ( x, y ) for all x, y ∈ X , intuitionistic fuzzy sub-sets of X × Y , having the form R = {〈 ( x, y ), [ M RL ( x, y ), M RU ( x, y )], [ N RL ( x, y ), N RU ( x, y )]〉} Y and N R = [0,0] for all x ∈ X . for all x ∈ X , y ∈ Y , (4) T4 : R is reflexive of type-4 if M RU ( x, x) ∧ where M RU ( y, y ) ≥ M RU ( x, y ), N RU ( x, x) ∨ Proposition 2: If A and B are be two GIVIFSs on X then obviously A ×1 B and A ×2 B are GIVIFSs
(1)
M RL ( x, y ) : X × Y → [0,1], M RU ( x, y ) : X × Y → [0,1] ,
N RU ( y, y ) ≤ N RU ( x, y ) for all x, y ∈ X . Definition 8: Let R ∈ GR( X × X ) , then and 0 ≤ M RU ( x) + N RU ( x) ≤ 1.5 . ' The collection of all GIVIFRs on X × Y is denoted (1) T1 : R is irreflexive of type-1 if M R = [0, 0], N R = [1,1] . by GR( X × Y ). Definition 5: Let R be a GIVIFR on X × Y , then we (2) T ' : R is irreflexive of type-2 if M = [0, 0] R 2 define the inverse of R , denoted by R −1 , where for all x ∈ X and for x,y ∈ X, N ( x, x) ∨ M RL ( x, y ) : X × Y → [0,1], M RU ( x, y ) : X × Y → [0,1] ,
M
R −1
( x, y ) = M R ( y, x) and N
R −1
RU
( x, y ) = N R ( y, x).
Definition 6: Let R ∈ GR( X × Y ) and P ∈ GR(Y × Z ) . Then we define two composite relations on X × Z , denoted by P R and P ∗ R where P R = {〈[ M ( P R ) L , M ( P R )U ],[ N ( P R ) L , N ( P R )U ]〉 : x ∈ X , z ∈ Z } and P ∗ R = {〈[M(P∗R)L , M(P∗R)U ],[N(P∗R)L , N(P∗R)U ]〉 : x ∈ X , z ∈Z}, and
N RU ( y, y ) ≥ 0.5 ∨ N RU ( x, y ). (3)
T3' : R is irreflexive of type-3 if M RU ( x, x) ∨ M RU ( y , y ) ≤ M RU ( x, y ) for all x, y ∈ X , N R = [1,1] for all x ∈ X .
(4)
T4' : R is irreflexive of type-4 if M RU ( x, x) ∨ M RU ( y, y ) ≤ M RU ( x, y ) and N RU ( x, x) ∧ N RU ( y, y ) ≥ N RU ( x, y ) for all x, y ∈ X .
International Journal of Fuzzy Systems, Vol. 14, No. 2, June 2012
196
The following results follows from definition. Theorem 1: (i) Reflexive (irreflexive) of type-1 ⇒ Reflexive (irreflexive) of type-2,3 and 4. (ii) Reflexive (irreflexive) of type-2 ⇒ Reflexive (irreflexive) of type-4. (iii) Reflexive (irreflexive) of type-3 ⇒ Reflexive (irreflexive) of type-4. Example 3: Let R be a reflexive of type T1 , where ⎡ [1,1],[0,0] ⎢ R = ⎢⎢ [0.1,0.3],[0.4,0.6] ⎢ [0.0,0.1],[0.6,0.8] ⎢⎣
[0.2,0.4],[0.4,0.6] [1,1],[0,0] [0.1,0.4],[0.6,0.8]
[0.5,0.7],[0.1,0.3] [0.5,0.8],[0.2,0.4] [1,1],[0,0]
⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦
Then T2 , T3 and T4 are obvious for R . Remark 1: It can easily be shown by constructing examples that reflexive (irreflexive) of type-4 ⇒ / reflexive reflexive (irreflexive) of type-3 ⇒ / (irreflexive) of type-2 ⇒ / reflexive (irreflexive) of type-1. Note 1: Reflexivity of type-4 which ⇒ / reflexive (irreflexive) of type-3 ⇒ / reflexive (irreflexive) of type-2 ⇒ / reflexive (irreflexive) of type-1 then it is obviously symmetric. Example 4: Let R be a reflexive of type-4, where ⎡ [0.1,0.3],[0.4,0.6] ⎢ R = ⎢⎢ [0.1,0.2],[0.4,0.8] ⎢ [0.0,0.1],[0.5,0.7] ⎣⎢
[0.1,0.2],[0.4,0.8] [0.1,0.4],[0.6,0.7] [0.1,0.3],[0.5,0.8]
[0.0,0.1],[0.5,0.7] [0.1,0.3],[0.5,0.8] [0.3,0.6],[0.3,0.5]
⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥
Here R ⇒ / reflexive (irreflexive) of type-3 ⇒ / reflexive (irreflexive) of type-2 ⇒ reflexive / (irreflexive) of type-1 and R is symmetric. Theorem 2: (i) If R ∈ GR( X × X ) is reflexive of any type then R ⊆ R R . (ii) If R ∈ GR( X × X ) is irreflexive of any type then R∗R ⊆ R . Proof: (i) Here we only prove for the upper limits of membership values and non-membership values. The prove for lower limits of membership values and non-membership values are similar. M ( R R )U ( x, z ) = max [min {M RU ( x, y ), M RU ( y, z )}] y
x
= max {min {M RU ( x, x ), M RU ( x, z )}, x
y
= min[ N RU ( x, z ),{min{max ( N RU ( x, y ), N RU ( y , z ))}}], y≠x
y
[ for any type of reflexivity of R ≤ N RU ( x , z ). ] Hence R ⊆ R R . (ii) Proof is similar to above. Now, we give an example, let R be a GIVIFR on X × Y i.e., R ∈ GR( X , Y ) which satisfy the property R ⊆ R R ( R ∗ R ⊆ R) though R is not any special type of reflexive(irreflexive). Example 5: Let R be a be a GIVIFR on X × Y i.e., R ∈ GR( X , Y ) , where ⎡ [0.2,0.3],[0.1,0.3] ⎢ R = ⎢⎢ [0.5,0.6],[0.1,0.3] ⎢ [0.2,0.3],[0.3,0.5] ⎣⎢
x
[ for any type of reflexivity of R ≥ M RU ( x , z ). ] N ( R R )U ( x, z ) = min [ max {N RU ( x, y ), N RU ( y, z )}] y
y
{min {max ( N RU ( x, y ), N RU ( y , z ))}}}
[0.2,0.3],[0.3,0.6] [0.3,0.6],[0.1,0.3] [0.2,0.3],[0.4,0.6]
⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥
(ii) If R1 and R2 is reflexive (irreflexive) of a particular type, then R1 ∩ R2 ( R1 ∪ R2 ) is reflexive (irreflexive) of same type. (iii) If R1 and R2 is reflexive (irreflexive) of same type then R1 ∪ R2 ( R1 ∩ R2 ) is reflexive (irreflexive) of same type. (iv) If R1 is reflexive (irreflexive) of type-1 and R2 be reflexive of any type then R1 ∪ R2 ( R1 ∩ R2 ) is reflexive (irreflexive) of type-1. Proof: (i) Here we only prove for the upper limits of membership values and non-membership values. The proves for lower limits of membership values and non-membership values are similar. Let R be reflexive of any type, then M ( R R )U ( x, x) = max {min {M RU ( x, y ), M RU ( y, x)}} x
= max {min {M RU ( x, x), M RU ( x, z )}, x
{max {min ( M RU ( x, y ), M RU ( y, z ))}}} y≠ x
x
= min [ M RU ( x, x),{max {min ( M RU ( x, y ), M RU ( y, z ))}}], x
x
= min {max{N RU ( x, x ), N RU ( x, z )},
[0.2,0.3],[0.3,0.6] ⎤ ⎥ [0.3,0.6],[0.1,0.3] ⎥⎥ [0.2,0.3],[0.3,0.6] ⎥⎦⎥
This shows that R ∗ R ⊆ R , though R is not any special type of irreflexive. Theorem 3: Let R, R1 , R2 ∈ GR ( X × X ) , then (i) If R is reflexive (irreflexive) of any type, then R R( R ∗ R) is reflexive (irreflexive) of same type.
y
y≠ x
x
[0.1,0.4],[0.3,0.5] [0.3,0.6],[0.2,0.3] [0.2,0.4],[0.3,0.5]
y
= min[ M RU ( x, z ),{max{min ( M RU ( x, y ), M RU ( y , z ))}}]
y≠ x
[0.4,0.7],[0.2,0.3] [0.2,0.4],[0.3,0.5]
⎡ [0.2,0.3],[0.4,0.5] ⎢ R ∗ R = ⎢⎢ [0.3,0.6],[0.1,0.3] ⎢ [0.2,0.3],[0.3,0.5] ⎣⎢
x
x
[0.4,0.6],[0.3,0.5]
and
{max {min ( M RU ( x, y ), M RU ( y , z ))}}} y≠ x
x
y≠x
x
= M RU ( x, x), for all x ∈ X .
(1)
N ( R R )U ( x, x) = N RU , for all x ∈ X .
(2)
Similarly If R is reflexive of type-3 or type-4, then we have, for
M. Bhowmik et al.: Some Results on Generalized Interval-Valued Intuitionistic Fuzzy Sets
y ≠ x.
197
Let P = R1 ∩ R2 , then for all x ∈ X
M (R
R )U ( x, y ) = max {min {M RU ( x, z ), M RU ( z , y )}}
M PU ( x, x) = min{M R U ( x, x), M R U ( x, x)}.
x
y
x
y
1
2
≤ min{min{M R U ( x, x), M R U ( y, y )},
x
1
{ max {min ( M RU ( x, z ), M RU ( z , y ))}}}
1
min{M R U ( x, x), M R U ( y, y )}}
x
2
= max [ M RU ( x, y ),{min ( M RU ( x, y ), M RU ( y, y ))},
2
= min{min{M R U ( x, x), M R U ( x, x)},
x
y
2
M PU ( x, y ) = min{M R U ( x, y ), M R U ( x, y )}
min {M RU ( x, y ), M RU ( y, y )}, z ≠ x, z ≠ y
1
Now for y ≠ x ,
= max {min {M RU ( x, x), M RU ( x, z )},
1
2
{ max {min ( M RU ( x, z ), M RU ( z , y ))}}],
min{M R U ( y, y ), M R U ( y, y )}}
≤ min {M RU ( x, x), M RU ( y, y )},
= min{M PU ( x, x), M PU ( y, y )}
z≠ x,z≠ y
x
1
x
For reflexive of type-3, M PU ( x, x) = min{M R U ( x, x), M R U ( x, x)} ≥ 0.5
since min {M RU ( x, x), M RU ( y, y )} ≥ M RU ( x, y ), x
= min{M ( R R )U ( x, x), M ( R R )U ( y, y )} by (1).
(3)
x
R )U
( x, x) = M R ( x, x) ≥ 0.5 ,
similarly M ( R
R )U
( y, y ) ≥ 0.5 .
So, min{M ( R
R )U
( x, x), M ( R
R )U
1
2
[since M R U ( x, x), and M R U ( x, x) ≥ 0.5 ] 1
If R is reflexive of type-3, then
M (R
2
( y, y )} ≥ 0.5 .
Therefore,
max{0.5, M(R R)U (x, y)}≤ min{M(R R)U (x, x), M(R R)U (y, y)}
2
Similarly, M PU ( y, y ) ≥ 0.5 . So, min{M PU ( x, x), M PU ( y, y )} ≥ 0.5 . Therefore, max{0.5, M PU ( x, y )} ≤ min{M PU ( x, x), M PU ( y, y )} Similarly, we can establish the result for N PU . Hence, from above P = R1 ∩ R2 is reflexive of the
(4) If R is reflexive of type-2 or type-4, then we have, for y ≠ x. N ( R R )U ( x, y ) = min {max{N RU ( x, z ), N RU ( z , y )}}
same type as that of R1 , R2 . The proof of irreflexivity is similar to that for reflexivity. (iii) The result is obvious for type-1 and type-2. Let P = R1 ∪ R2 , then ∀x ∈ X
= min{max{ N RU ( x, x), N RU ( x, z )}, max{N RU ( x, y ), N RU ( y, y )}, = min{max{ N RU ( x, x), N RU ( x, z )}, max{N RU ( x, y ), N RU ( y, y )}, { min {max( N RU ( x, z ), N RU ( z , y ))}}}
M PU ( x, x) = max{M R U ( x, x), M R U ( x, x)},
z
z ≠ x, z ≠ y
1
Now for y ≠ x
2
M PU ( x, y ) = max{M R U ( x, y ), M R U ( x, y )} 1
2
≤ max{min{M R U ( x, x), M R U ( y, y )}, 1
1
min{M R U ( x, x), M R U ( y, y )}} 2
2
= min[ N RU ( x, y ),{max ( N RU ( x, y ), N RU ( y, y ))},
= min[{max{min{M R U ( x, x), M R U ( y, y )}, M R ( x, x)},
{ min {max( N RU ( x, z ), N RU ( z , y ))}}],
{max{min{M R U ( x, x), M R U ( y, y )}, M R ( y, y )}]
z ≠ x, z ≠ y
1
1
2
2
2
2
≥ max{N RU ( x, x), N RU ( y, y )},
= min[min{max{M R U ( x, x), M R U ( x, x)},
[since max{N RU ( x, x), N RU ( y, y )} ≥ N RU ( x, y ), ∀x, y ∈ X ]
max{M R U ( y, y ), M R U ( x, x)}},
= max{N ( R R )U ( x, x), N ( R R )U ( y, y )} by (2). (5) Hence by equation (1) to (5), R R is reflexive of the type as that of R . The proof of irreflexive is similar to that for reflexivity. (ii) Here also we only prove for the upper limits of membership values and non-membership values and for the lower limits of membership values and nonmembership values are similar.
1
1
2
2
min{max{M R U ( x, x), M R U ( y, y )}, 1
2
max{M R U ( y, y ), M R U ( y, y )}}] 1
2
= min{max{M R U ( x, x), M R U ( x, x)}, 1
2
max{M R U ( y, y ), M R U ( x, x)}} 1
2
For reflexive of type-3, M PU ( x, x) = max{M R U ( x, x), M R U ( x, x)} ≥ 0.5 1
2
International Journal of Fuzzy Systems, Vol. 14, No. 2, June 2012
198
(since M R U ( x, x), M R U ( x, x) ≥ 0.5 ) 1
and Y
2
respectively, If R1 , R2 be two subsets of
ˆ R are not R ∈ GR( X × Y ) . Then R1 ⊕ R2 , R1 ⊕ 2 subsets of R .
Similarly, M PU ( y, y ) ≥ 0.5 . So, min{M PU ( x, x), M PU ( y, y )} ≥ 0.5 . Therefore, max{0.5, M PU ( x, y )} ≤ min{M PU ( x, x ), M PU ( y, y )} .
We shown it by means of example. Example 7: Let A and B be two GIVISs A = {〈 a, [0.1,0.3], [0.6,0.7]〉, 〈b, [0.4,0.5], [0.2,0.6]〉,
〈c , [0.3,0.4], [0.3,0.6] 〉}.
Similarly, we can establish the result for N PU . Hence, from above P = R1 ∪ R2 is reflexive of the same type as that of R1 , R2 . The proof of irreflexivity is similar to that for reflexivity. (iv) The proof is similar to that of above case. Example 6: Let R1 be a reflexive of type-1 and R2 be another reflexive of any type where ⎡ [1,1],[0,0] ⎢ ⎢ R1 = ⎢ [0.1,0.3],[0.4,0.6] ⎢ [0.0,0.1],[0.6,0.8] ⎣⎢
[0.2,0.4],[0.4,0.6] [1,1],[0,0]
[0.5,0.7],[0.1,0.3] [0.5,0.8],[0.2,0.4]
[0.1,0.4],[0.6,0.8]
[1,1],[0,0]
⎡ [1,0.4],[0,0.5] ⎢ R2 = ⎢⎢ [0.1,0.5],[0.4,0.4] ⎢ [0.1,0.3],[0.3,0.5] ⎣⎢
[0.2,0.5],[0.4,0.5] [1,0.5],[0,0.3] [0.1,0.4],[0.6,0.5]
[0.5,0.6],[0.1,0.5] [0.5,0.7],[0.2,0.4] [1,0.2],[0,0.4]
⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥
Then, R1 ∪ R2 = ⎡ [1,1],[0,0] ⎢ ⎢ [0.1,0.5],[0.4,0.4] ⎢ ⎢ [0.0,0.1],[0.3,0.5] ⎣⎢
[0.2,0.5],[0.4,0.5] [1,1],[0,0]
[0.5,0.7],[0.1,0.3] [0.5,0.8],[0.2,0.4]
[0.1,0.4],[0.6,0.5]
[1,1],[0,0]
⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥
This shows that R1 ∪ R2 is reflexive of type-1.
B = {〈 x, [0.3,0.6], [0.1,0.4]〉, 〈 y, [0.7,0.9], [0.2,0.5]〉,
〈 z , [0.3,0.4], [0.3,0.6] 〉}. Then, A× B x
y
a ⎡ [0.1,0.3],[0.6,0.7] ⎢ = b ⎢⎢ [0.3,0.5],[0.2,0.6] c ⎢⎢⎣ [0.3,0.5],[0.1,0.4]
z
[0.1,0.3],[0.6,0.7] [0.4,0.5],[0.2,0.6] [0.3,0.5],[0.2,0.5]
[0.1,0.3],[0.6,0.7] [0.3,0.4],[0.3,0.6] [0.3,0.4],[0.3,0.6]
⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦
Let, x
y
a ⎡ [0.1,0.2],[0.6,0.8] ⎢ R1 = b ⎢⎢ [0.2,0.3],[0.3,0.6] c ⎢⎢⎣ [0.3,0.4],[0.2,0.5]
z
[0.1,0.3],[0.7,0.9] [0.3,0.4],[0.3,0.6] [0.2,0.4],[0.2,0.5]
⎤ ⎥ ⎥ ⎥ [0.1,0.2],[0.3,0.7] ⎥⎥⎦
[0.1,0.2],[0.6,0.7] [0.1,0.4],[0.4,0.6]
and x ⎡ a [0.1,0.3],[0.6,0.9] ⎢ R2 = b ⎢⎢ [0.2,0.4],[0.2,0.7] c ⎢⎣⎢ [0.3,0.5],[0.2,0.6]
Here, N( R ⊕R 1
2 )U
( a, y ) =
y
z
[0.1,0.2],[0.7,0.8] [0.4,0.5],[0.3,0.7] [0.2,0.5],[0.3,0.7]
[0.1,0.2],[0.6,0.8] [0.1,0.3],[0.4,0.7] [0.2,0.3],[0.4,0.8]
N R U ( a, y ) + N R U ( a, y ) 1
2
2.{N R U (a, y ).N R U (a, y ) + 1} 1
⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥
= 0.31,
2
but N ( A× B )U (a, y ) = 0.7 .
5. Some more results on GIVIFSs
Hence N ( R ⊕ R 1
Definition 9: For any two GIVIFSs A and B on X , we define two operations as follows: MAU (x) + MBU (x) MAL(x) + MBL(x) (1) A⊕B ={[ 〈 , ], 2(MAL(x).MBL(x) +1) 2(MAU (x).MBU (x) +1)
N AU ( x) + N BU ( x) N AL ( x) + N BL ( x) , ]〉}. 2(N AL ( x).N BL ( x) + 1) 2(N AU ( x).N BU ( x) + 1) MAU (x) + MBU (x) MAL(x) + MBL(x) ˆ B ={[ A⊕ 〈 , ], 2(MAL(x).MBL(x) +1) 2(MAU (x).MBU (x) +1) NAU (x) + NBU (x) NAL(x) + NBL(x) , ]〉}. [ 2(NAL(x) + NBL(x) +1) 2(NAU (x) + NBU (x) +1) [
(2)
Note 2: As for a, b ∈ [0,1] ,
( a + b) ≤ 0.5 and 2(ab + 1)
( a + b) ˆ B are ≤ 0.5 , therefore A ⊕ B and A ⊕ 2(a + b + 1) also GIVIFSs. Remark 2: For any two GIVIFSs A and B on X
2 )U
(a, y ) ≥/ N ( A× B )U (a, y )
i.e., R1 ⊕ R2 ⊆ / ( A × B) . And, N( R1⊕ˆ R2 )U (a, y) =
N R1U (a, y) + N R2U (a, y) 2.{N R1U (a, y) + N R2U (a, y) + 1}
= 0.49,
but N ( A× B )U (a, y ) = 0.7 . Hence N ( R1⊕ˆ R2 )U (a, y ) ≥/ N ( A×B )U (a, y )
ˆ R ⊄ ( A × B) . i.e., R1 ⊕ 2 Definition 10: For any two GIVIFSs A and B on X and Y respectively and let R ∈ GR( X × Y ) . Then we define ∧ R = {〈[min { N RL ( x , y ), M ( A× B ) L ( x , y )},
min {N RU ( x, y ), M ( A× B )U ( x, y )}], [max {M RL ( x, y ), N ( A×B ) L ( x, y )}, {max M RU ( x, y ), N ( A× B )U ( x, y )}]〉}.
M. Bhowmik et al.: Some Results on Generalized Interval-Valued Intuitionistic Fuzzy Sets
for all x ∈ A and y ∈ B. Theorem 4: For any two GIVIFSs A and B on X and Y respectively and let R ∈ GR( X × Y ) , then
ˆ R ⊆ Rˆ . Proof: We know R = {〈[ M RL , M RU ],[ N RL , N RU ]〉} . Then Rˆ = {〈[ M ˆ , M ˆ ],[ M ˆ , M ˆ ]〉} , where RL
RU
RL
RU
NRL ˆ = max{MRL , N( A×B)L}, N∧ = max{MRU , N( A×B)U } RU
y
l a ⎡ [0.1,0.3],[0.6,0.7] ⎢ l R1 = b ⎢⎢ [0.2,0.6],[0.3,0.5] c ⎢⎣⎢ [0.3,0.4],[0.2,0.5] ˆ That is R1 ⊆ Rˆ1.
z ⎤ ⎥ ⎥ ⎥ [0.3,0.4],[0.3,0.6] ⎥⎦⎥
[0.1,0.3],[0.6,0.7] [0.3,0.5],[0.3,0.6]
[0.1,0.3],[0.6,0.7] [0.3,0.4],[0.3,0.6]
[0.2,0.4],[0.2,0.5]
(2) ( R1−1 ) −1 = R1
and
(3) ( R1 ∪ R2 ) −1 = R1−1 ∪ R2−1 , ( R1 ∩ R2 ) −1 = R1−1 ∩ R2−1 Proof (1):
Mˆˆ =min{NRL ˆ ,M(A×B)L}, Mˆˆ =min{NRU ˆ ,M(A×BU ) }, RU
N ˆˆ =max{MRL ˆ , N( A×B)L}, N ˆˆ =max{MRU ˆ , N( A×B)U}. RL
x
Theorem 5: Let A and B be two GIVIFSs on X and Y respectively and R ∈ GR( X × Y ) . If R1 , R2 be two subsets of R . Then (1) R1 ⊆ R2 ⇔ R1−1 ⊆ R2−1
M RL ˆ = min{ N RL , M ( A× B ) L }, M RL ˆ = min{ N RU , M ( A× B )U },
RL
199
R1 ⊆ R2 ⇔ {M R L ( x, y ) ≤ M R L ( x, y ),
RU
1
We have to show,
2
M RL ≤ min {max ( M RL , N ( A×B ) L ), M ( A× B ) L },
(6)
M R U ( x, y ) ≤ M R U ( x, y ) and
M RU ≤ min {max ( M RU , N ( A× B )U ), M ( A× B )U },
(7)
N R L ( x, y ) ≥ N R L ( x, y ), N R U ( x, y ) ≥ N R U ( x, y )}
1
2
1
2
1
2
N RL ≥ max {min ( N RL , M ( A× B ) L ), N ( A× B ) L },
(8)
⇔ {M
N RU ≥ max {min (N RU , M ( A× B )U ), N ( A× B )U },
(9)
N R−1L ( y, x) ≥ N R−1L ( y, x), N R−1U ( y, x) ≥ N R−1U ( y, x)}
ˆˆ RL
= min { N
ˆˆ , M ( A× B ) L } RL
= min {max ( M RL , N ( A×B ) L ), M ( A× B ) L }
(10)
By definition, M RL ≤ M ( A× B ) L , M RU ≤ M ( A× B )U
and N RL ≥ N ( A×B ) L , N RU ≥ N ( A×B )U
Case
1:
Let
max{M RL , N ( A× B ) L } = M RL
then,
min{M RL , M ( A× B ) L } = M RL i.e., M RL = M ˆˆ . then
( y, x), M
2
−1 1
M RL < M ˆˆ
RL
since M RL ≤ M ( A× B ) L .
a ⎡ [0.0,0.1],[0.6,0.9] ⎢ −1 R1 = b ⎢⎢ [0.1,0.2],[0.7,0.9] c ⎢⎢⎣ [0.1,0.1],[0.7,0.8]
a ⎡ [0.1,0.2],[0.6,0.8] ⎢ R2−1 = b ⎢⎢ [0.1,0.3],[0.7,0.9] c ⎢⎢⎣ [0.1,0.2],[0.6,0.7]
and so
∧ R1 = R1 . Again,
[0.1,0.3],[0.6,0.7] [0.3,0.5],[0.3,0.6] [0.2,0.5],[0.2,0.4]
2
x a ⎡ [0.0,0.1],[0.6,0.9] ⎢ R1 = b ⎢⎢ [0.1,0.2],[0.4,0.6] c ⎢⎣⎢ [0.3,0.3],[0.3,0.6] x a ⎡ [0.1,0.2],[0.6,0.8] ⎢ R2 = b ⎢⎢ [0.2,0.3],[0.3,0.6] c ⎢⎢⎣ [0.3,0.4],[0.2,0.5]
(10) reduces to case 1. ˆˆ . Similarly equations (7), (8), (9) can Hence M RL ≤ M RL be proved. Example 8: From the Example 7, we get a ⎡ [0.1,0.3],[0.6,0.7] ⎢ l R1 = b ⎢⎢ [0.3,0.5],[0.2,0.6] c ⎢⎣⎢ [0.2,0.5],[0.3,0.4]
1
( y, x) and
Other proofs are straightforward. Example 9: For the example 7, let R1 and R2 be
and
y
( y, x) ≤ M
−1 R2U
⇔R ⊆R
When min {N ( A× B ) L , M ( A× B ) L } = N ( A× B ) L then equation
x
−1 R1U
−1 2 .
x
Case 2: When max{M RL , N ( A× B ) L } = N ( A× B ) L min{N ( A× B ) L , M ( A× B ) L } = M ( A× B ) L ,
−1 R2 L
y
z
[0.1,0.2],[0.7,0.9] [0.2,0.3],[0.3,0.7] [0.1,0.3],[0.3,0.6] y
[0.1,0.1],[0.7,0.8] [0.1,0.2],[0.5,0.8] [0.1,0.1],[0.4,0.7] z
⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥
[0.1,0.3],[0.7,0.9] [0.3,0.4],[0.3,0.6] 2 [0.1,0.4],[0.2,0.5]
[0.1,0.2],[0.6,0.7] [0.1,0.4],[0.4,0.6] [0.1,0.2],[0.3,0.7]
⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦
Now,
RL
and,
( y, x) ≤ M
1
We will prove only equation (6), and other proofs are similar. M
−1 R1L
z [0.1,0.3],[0.6,0.7] ⎤ ⎥ [0.3,0.4],[0.3,0.6] ⎥⎥ [0.3,0.4],[0.3,0.6] ⎥⎦⎥
x
y
z
[0.1,0.2],[0.4,0.6] [0.2,0.3],[0.3,0.7] [0.1,0.2],[0.5,0.8]
[0.3,0.3],[0.3,0.6] [0.1,0.3],[0.3,0.6] [0.1,0.1],[0.4,0.7]
y
⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦
z
[0.2,0.3],[0.3,0.6] [0.3,0.4],[0.3,0.6] 2 [0.1,0.4],[0.4,0.6]
[0.3,0.4],[0.2,0.5] [0.2,0.4],[0.2,0.5] [0.1,0.2],[0.3,0.7]
y
z
[0.1,0.3],[0.7,0.9] [0.4,0.5],[0.3,0.6] 2 [0.2,0.5],[0.2,0.5]
[0.1,0.2],[0.6,0.7] [0.1,0.4],[0.4,0.6] [0.2,0.3],[0.3,0.7]
⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦
So, R1−1 ⊆ R2−1 . R1 ∪ R2 = x ⎡ a [0.1,0.2],[0.6,0.8] ⎢ b ⎢⎢ [0.2,0.4],[0.2,0.6] c ⎢⎢⎣ [0.3,0.5],[0.2,0.5]
so,
⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦
International Journal of Fuzzy Systems, Vol. 14, No. 2, June 2012
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'′ '′ K RL = min M RL ( x, y ); K RU = min M RU ( x, y )
( R1 ∪ R2 )−1 = x
y
a ⎡ [0.1,0.3],[0.6,0.8] ⎢ b ⎢⎢ [0.1,0.3],[0.7,0.8] c ⎢⎢⎣ [0.1,0.2],[0.6,0.7] R1−1 ∪ R2−1 = x ⎡ a [0.1,0.3],[0.6,0.8] ⎢ b ⎢⎢ [0.1,0.3],[0.7,0.8] c ⎢⎢⎣ [0.1,0.2],[0.6,0.7]
( x , y )∈ X × X
z [0.3,0.5],[0.2,0.5] ⎤ ⎥ [0.2,0.5],[0.2,0.5] ⎥⎥ [0.2,0.3],[0.3,0.7] ⎥⎥⎦
[0.2,0.4],[0.2,0.6] [0.4,0.5],[0.3,0.6] 2 [0.1,0.4],[0.4,0.6] y
z
[0.2,0.4],[0.2,0.6] [0.4,0.5],[0.3,0.6] 2 [0.1,0.4],[0.4,0.6]
[0.3,0.5],[0.2,0.5] [0.2,0.5],[0.2,0.5] [0.2,0.3],[0.3,0.7]
⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦
Hence, ( R1 ∪ R2 ) −1 = R1−1 ∪ R2−1 .
6. C and I two operators on GIVIFSs
'′ RL
L
( x , y )∈ X × X
'′ RU
= max N RL ( x, y ); L ( x , y )∈ X × X
= max N RU ( x, y ). ( x , y )∈ X × X
Note 3: It is obvious that C (R ) may not be a subset of R but I (R) is always a subset of R . Theorem 7: For A, B ∈ GIVIFSs and R ∈ GR( A × B) , then (i) C (R ) is subset of C ( A) × C ( B ) , (ii) I (R) is subset of I ( A) × I ( B) . Proof: (i) We know from definition ' ' C ( R) = {〈 ( x, y ), [ K RL , K RU ], [ L'RL , L'RU ]〉}, where ' K RL = max M RL ( x, y ), L'RL = min N RL ( x, y )
K
I ( A) = {〈[ K L'′ , KU'′ ], [ L'L′ , LU'′ ]〉} , where
Since,
K L' = max M AL ( x) and KU' = max M AL ( x), x∈ X
x∈ X
L'L = min N AL ( x) and LU' = min N AL ( x), x∈ X
'′ L
x∈ X
'′ U
K = min M AL ( x) and K = min M AU ( x), '′ L
x∈ X
x∈ X
'′ U
L = max N AL ( x) and L = max N AU ( x).
( x , y )∈ X × X
( x , y )∈ X × X
Definition 11: For every GIVIFSs, A , we define the following subsets C ( A) = {〈[ K L' , KU' ], [ L'L , LU' ]〉} and
' RU
' RU
= max M RU ( x, y ), L ( x , y )∈ X × X
= min N RU ( x, y ) ( x , y )∈ X × X
M RL ( x, y ) ≤ M ( A× B ) L ( x, y ) = min{M A ( x), M B ( y )} So, M RL ( x, y ) ≤ M A ( x) and M RL ( x, y ) ≤ M B ( y ) ⇒ max M RL ( x, y ) ≤ max M A ( x) ( x, y )
x
and maxM RL ( x, y ) ≤ maxM B ( y ) ( x, y )
y
' ⇒ K ≤ K A and K RL ≤ KB Theorem 6: If A is a GIVIFS on X then C (A) and ' ⇒ K ≤ K A and K RL ≤ K B I ( A) are also GIVIFSs on X . ' ⇒ K RL ≤ min{K A , K B } Proof: We have to show KU' ∧ LU' ≤ 0.5 and ' ≥ max{LA , LB } . Similarly, we can prove K RL '′ '′ KU ∧ LU ≤ 0.5 . So, C (R) is a subset of C ( A) × C ( B). ' Clearly, if KU > 0.5 , then there exist x1 ∈ X such Theorem 8: Let A nd B be two GIVIFSs and that M AU ( x1 ) > 0.5, as otherwise ∨ x∈X M AU (x) can R ∈ GR ( A × B ) , then (i) C ( A × B ) = C ( A) × C ( B ) . not exceed 0.5 . (ii) I ( A × B) = I ( A) × I ( B) . But then N AU ( x1 ) ≤ 0.5 and hence
x∈ X
LU' = min M AU ( x) ≤ M AU ( x1 ) ≤ 0.5 x∈ X
' U
'′ U
'′ U
Similarly, we can prove K ∧ L ≤ 0.5 . Hence C ( A) and I (A) are GIVIFSs on X . Definition 12: Let A and B be two GIVIFSs on X and Y respectively and R ∈ GR( A × B) , we use the ' ' following subsets C ( R ) = {〈[ K RL , K RU ], [ L'RL , L'RU ]〉} ′
′
′
′
' ' and I ( R ) = {〈[ K RL , K RU ], [ L'RL , L'RU ]〉} , where, ' ' K RL = max M RL ( x, y ); K RU = max M RU ( x, y ) ( x , y )∈ X × X
' RL
L
( x , y )∈ X × X
' RU
= min N RL ( x, y ); L ( x , y )∈ X × X
Proof: The proofs are straightforward. Example 10: Let, x
so, K ∧ L ≤ 0.5. ' U
' RL ' RL
x∈ X
= min N RU ( x, y ) ( x , y )∈ X × X
y
a ⎡ [0.1,0.2],[0.7,0.8] ⎢ R1 = b ⎢⎢ [0.2,0.4],[0.4,0.5] c ⎢⎢⎣ [0.3,0.4],[0.2,0.6]
[0.1,0.3],[0.7,0.9] [0.3,0.4],[0.3,0.7] [0.2,0.5],[0.3,0.6]
z ⎤ ⎥ ⎥ ⎥ [0.1,0.2],[0.4,0.7] ⎥⎥⎦ [0.1,0.2],[0.6,0.7] [0.1,0.3],[0.4,0.6]
C ( A) = 〈 a, [0.4,0.5], [0.2,0.4]〉, 〈b, [0.4,0.5], [0.2,0.4]〉,
〈c, [0.4,0.5], [0.2,0.4]〉. C ( B) = 〈 x, [0.7,0.9],[0.1,0.4]〉, 〈 y, [0.7,0.9],[0.1,0.4]〉,
〈 z , [0.7,0.9],[0.1,0.4]〉. C ( A × B ) = {〈 (a1 , x1 )[0.4,0.5], [0.2,0.4]〉 : (a1 , x1 ) ∈ X × X }. C ( R ) = {〈 (a1 , x1 )[0.3,0.4], [0.2,0.5]〉 : for all (a1 , x1 ) ∈ X × X }.
M. Bhowmik et al.: Some Results on Generalized Interval-Valued Intuitionistic Fuzzy Sets
since
M C ( R ) L = 0.3 and M C ( R )U = 0.4 but M ( A× B ) L (a, x) = 0.1 and M ( A× B )U (a, x) = 0.3 i.e., M C ( R ) L ≤/ M ( A× B ) L and M C ( R )U ≤/ M ( A× B ) L . So, C ( R ) ⊆ / A× B . Hence C (R ) may not be a subset of R . I ( A) = 〈 a, [0.1,0.3], [0.6,0.7]〉, 〈b, [0.1,0.3], [0.6,0.7]〉,
〈c, [0.1,0.3], [0.6,0.7]〉. I ( B) = 〈 x, [0.3,0.4],[0.3,0.6]〉, 〈 y, [0.3,0.4],[0.3,0.6]〉 ,
201
⎧1 ⎪2 ,if MRiL =1 for i =1,2,…, n. ⎪⎪ n−1 MRL = ⎨ [(−1)k+1 ∑ ∑ M(Ri1L).M(Ri2L) M(RikL) ] ⎪ k=1 i1
